PROBABILITY OPERA PROBLEM IT LOOKS SIMPLE BUT IT IS A CHAL

PROBABILITY - OPERA PROBLEM - IT LOOKS SIMPLE BUT IT IS A CHALLENGE (this question seems to be easy but there are two big subtles things that are very trick).

ATTENTION: If any theorem, definition, lemma or formula is used in the answer, it should be stated and explained why it is been used. Even though the problem may look simple, THIS IS A PROBABILITY QUESTION (Solution needs to be shown STEP BY STEP with MATHEMATICAL JUSTIFICATION - FORMALLY). Solution wihtout detailed work won\'t be marked. Solution needs to show how and why. FORMAL ANSWER should be applied.

Solution

There are 5 people and they leave their coats and hats.

Thus there are 5 coats and 5 hats.

After Opera, they are given one coat and hat in random order.

Prob for one person to receive one coat and hat correct is P(correct coat)P(correct hat)

as these two events are independent.

Prob for correct coat out of 5 = 1/5 = 0.2

Prob for correct hat   = 0.2

p = Prob for a person to receive correct hat and coat = 0.2(0.2) = 0.04

Prob (2 persons receiving correct by chance) = 1/5(1/5)(1/4)(1/4) = 0.0025

P(x=2) = Prob for 2 persons receiving correct coat and hat and other 3 not receiving correct coat and hat together.

If 2 persons have taken correct coat and hat, remaining will be 3 coats and 3 hats to be given to 3 persons.

P(each person not receiving their correct coat and hat together)

= 1-Prob of all 3 receiving correct coat and hat together

= 1-(1/3)(1/3) (1/2)(1/2)

= 1-1/36

= 35/36

=0.9722

Also 2 persons receiving correct can be in 5 C2 ways.

Hence required prob = 5C2(0.96)(0.0025)

=0.024